1. **State the problem:** We have a circle with equation $$(x - 5)^2 + y^2 = 25$$ and a point $$P(8,4)$$ on the circle. We need to find the y-coordinate of point $$Q$$ where the tangent at $$P$$ crosses the y-axis. 2. **Formula and rules:** The tangent to a circle at point $$P(x_1,y_1)$$ on the circle $$ (x - h)^2 + (y - k)^2 = r^2 $$ has the equation: $$ (x_1 - h)(x - h) + (y_1 - k)(y - k) = r^2 $$ Here, the circle center is $$ (h,k) = (5,0) $$ and radius $$ r = 5 $$. 3. **Apply the formula:** Substitute $$x_1=8$$, $$y_1=4$$, $$h=5$$, $$k=0$$, and $$r=5$$: $$ (8 - 5)(x - 5) + (4 - 0)(y - 0) = 25 $$ Simplify: $$ 3(x - 5) + 4y = 25 $$ 4. **Expand and simplify:** $$ 3x - 15 + 4y = 25 $$ $$ 3x + 4y = 40 $$ 5. **Find y-intercept (where tangent crosses y-axis):** At y-axis, $$x=0$$, so: $$ 3(0) + 4y = 40 $$ $$ 4y = 40 $$ $$ y = 10 $$ **Final answer:** The y-coordinate of point $$Q$$ is $$\boxed{10}$$.